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https://leetcode.com/problems/length-of-last-word/
Given a string s consists of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
For example,
Given s = "Hello World",
return 5.
调用String.sqilt(" ")来求解：
1 public class Solution { 2 public int lengthOfLastWord(String s) { 3 String[]t=s.split(" "); 4 if(t.length==0)return 0; 5 return t[t.length-1].length(); 6 } 7 }
自己实现：
1 public class Solution { 2 public static int lengthOfLastWord(String s) { 3 if(s.equals("")) return 0; 4 int len=s.length(); 5 int index=len; 6 while(index>0&&s.charAt(--index)==‘ ‘); 7 if(s.charAt(index)==‘ ‘) return 0; 8 int cout=1; 9 while(index>0&&s.charAt(--index)!=‘ ‘){cout++;} 10 return cout; 11 } 12 public static void main(String[]args){ 13 System.out.println(lengthOfLastWord(" ")); 14 } 15 }
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原文地址：http://www.cnblogs.com/qq1029579233/p/4480763.html
